Because 1 is an int, 32 bits, so (1 << 27)*27 overflows. Use 1ull. Regarding your comment, if x is a uint64_t, then 1 << x is still an int, but for the multiplication it would be cast to uint64_t, so there’d be no overflow. However, if x >= 31, 1 << x would be … Read more
In ECMAScript (Javascript) bitwise operations are always in 32-bit. Therefore 5799218898 is chopped into 32-bit which becomes 1504251602. This integer >> 13 gives 183624. In Python they are arbitrary-length integers. So there’s no problem. (And the numbers in Windows calculator are 64-bit, enough to fit 5799218898.) (And the correct answer should be 707912.)
I’d use: if ((value & (1L << x)) != 0) { // The bit was set } (You may be able to get away with fewer brackets, but I never remember the precedence of bitwise operations.)
Although your x is of type long int, the 1 is not. 1 is an int, so 1<<63 is indeed undefined. Try (static_cast<long int>(1) << 63), or 1L << 63 as suggested by Wojtek.
int x = (number >> (8*n)) & 0xff; where n is 0 for the first byte, 1 for the second byte, etc.
The statement: if (counter & (1<<j)) checks if the j-th bit of counter is set. In more detail, 1 << j uses shifting of 1 to generate a bit mask in which only the j-th bit is set. The & operator then masks out the j-bit of counter; if the result is not zero (which … Read more
You’re correct; it is used to truncate the value. The reason >> works is because it operates only on 32-bit integers, so the value is truncated. (It’s also commonly used in cases like these instead of Math.floor because bitwise operators have a low operator precedence, so you can avoid a mess of parentheses.) And since … Read more
Well, the official Java tutorial Bitwise and Bit Shift Operators covers the actual operations that are available in Java, and how to invoke them. If you’re wondering “what can I do with bit-shifting”, then that’s not Java specific, and since it’s a low-level technique I’m not aware of any list of “cool things you can” … Read more
Because in Java there are no unsigned datatypes, there are two types of right shifts: arithmetic shift >> and logical shift >>>. http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html Arithmetic shift >> will keep the sign bit. Unsigned shift >>> will not keep the sign bit (thus filling 0s). (images from Wikipedia) By the way, both arithmetic left shift and logical … Read more